Question

Consider a binary mixture of components \( A \) and \( B \) at temperature \( T \) and pressure \( P \). Let \( \bar{V}_A \) and \( \bar{V}_B \) be the partial molar volumes of \( A \) and \( B \), respectively. At a certain mole fraction of \( A \), \( x_A \): \[ \left( \frac{\partial \bar{V}_A}{\partial x_A} \right)_{T,P} = 22 \, \text{cm}^3 \, \text{mol}^{-1}, \quad \left( \frac{\partial \bar{V}_B}{\partial x_A} \right)_{T,P} = -18 \, \text{cm}^3 \, \text{mol}^{-1}. \] The value of \( x_A \), rounded off to 2 decimal places, is \_\_\_\_\_\_\_.

Hint:

For binary mixtures, apply the Gibbs-Duhem relation to relate partial molar properties and mole fractions.

Answer 1

Step 1: Use the Gibbs-Duhem relation. The Gibbs-Duhem relation for partial molar properties in a binary mixture is: \[ \left( \frac{\partial \bar{V}_A}{\partial x_A} \right)_{T,P} x_A + \left( \frac{\partial \bar{V}_B}{\partial x_A} \right)_{T,P} (1 - x_A) = 0. \] Step 2: Substitute the given values. \[ 22x_A + (-18)(1 - x_A) = 0. \] Simplify: \[ 22x_A - 18 + 18x_A = 0 \quad \Rightarrow \quad 40x_A = 18 \quad \Rightarrow \quad x_A = \frac{18}{40} = 0.45. \] Step 3: Conclusion. The mole fraction of \( A \) is \( x_A = 0.45 \).